We work in ZFC throughout. The following question was posed to me by a friend:
Can there exist cardinals $\kappa,\lambda$ such that $\lambda<\mathrm{cof}(\kappa)$ and $2^\lambda<\kappa<\kappa^\lambda$?
Originally I thought this should be easy, after all many minor variants are almost immediate: if we allow $\lambda=\mathrm{cof}(\kappa)$, then Konig's theorem implies $\kappa^\lambda>\kappa$ and we can easily arrange $2^\lambda$ to be small (either by choosing a model where $2^\lambda$ is small, or picking $\kappa$ large enough), and if we don't require $2^\lambda<\kappa$, then we can for instance let $\kappa=\aleph_{\omega_1}$ and take a model in which $2^{\aleph_0}$ is larger than that.
However, thinking about this problem further made me realize this is probably quite close to some unsolved problems in set theory. The smallest viable counterexample could be given by $\kappa=\aleph_{\omega_1}$ and $\lambda=\aleph_0$. It is not hard to see that for $\kappa^\lambda>\kappa$ to hold, we need $\mu^\lambda>\kappa$ for some $\mu<\kappa$ (this is because of the inequality $\lambda<\mathrm{cof}(\kappa)$). Therefore one idea to get the result is to pick a model in which $2^{\aleph_0}<\aleph_{\omega_1}$ and $\aleph_\omega^{\aleph_0}>\aleph_{\omega_1}$. I have no idea whether this is expected to be possible, but some results and conjectures in Shelah's PCF theory suggest the answer might be negative – specifically, if we ask for a stronger inequality $2^{\aleph_0}<\aleph_{\omega}$, then it is known $\aleph_\omega^{\aleph_0}<\aleph_{\omega_4}$, and it is conjectured that $<\aleph_{\omega_1}$ should also be true then.
Of course, those results are not strong enough to exclude it, and they show that our tools are not good enough to disprove that result. Is it then, against all odds, consistent with ZFC that such a pair of cardinals exists?
Some comments indicate the answer should be positive under large cardinal (consistency) assumptions. I am primarily interested in whether this is relatively consistent with ZFC alone, but if no such results are available I am happy to accept an answer conditional on higher consistency strength.
Best Answer
It is consistent that such a pair exists, see my paper Singular cofinality conjecture and a question of Gorelic.
To show that some large cardinals are needed, suppose for example $\lambda=\aleph_0 < \aleph_1=cf(\kappa)$ and $\kappa^\omega > \kappa > 2^\omega.$ Then for some $\mu < \kappa, \mu^\omega > \kappa,$ and without loss of generality $2^\omega < \mu$ and $cf(\mu)=\aleph_0$ (otherwise pick some $\mu'$ in the interval $(\mu, \kappa)$ as required). Thus SCH (the singular cardinals hypothesis) fails and hence we need some large cardinals. Indeed $\mu^\omega \geq \kappa^+ \geq \mu^{+\omega_1+1}$, so it sems a large cardinal $\theta$ with $o(\theta) \geq \theta^{+\omega_1+1}$ is needed.
Suppose in general $\lambda < cf(\kappa)$ and $2^\lambda < \kappa < \kappa^{\lambda}$. We may assume that $\lambda$ is the least cardinal with this property. Let me first show that $\lambda$ is regular. Otherwise let $(\lambda_i: i< cf(\lambda))$ be an increasing sequence of regular cardinals cofinal in $\lambda$. By the choice of $\lambda,$ for all $i, \kappa^{\lambda_i} \leq \kappa$ (as clearly for all $i, \lambda_i< cf(\kappa)$ and $2^{\lambda_i}<\kappa$), thus $\kappa^\lambda= \prod_{i<cf(\lambda)}\kappa^{\lambda_i} \leq \kappa^{cf(\lambda)} \leq \kappa^\lambda$. This shows that $\lambda'=cf(\lambda)$ also satisfies $\lambda' < cf(\kappa)$ and $2^{\lambda'} < \kappa < \kappa^{\lambda'}$, which contradicts the choice of $\lambda$ as such a minimal cardinal. Now the above argument works with essentially the same argument.